1. Conceptual Analysis

The problem asks for the acceleration of a small metal disc just as the glue holding it to a porcelain ball fails. We are dealing with electrostatic pressure on a charged surface.

• System: A large porcelain (insulating) ball of radius \( R \) with a small metal disc of radius \( r \) and mass \( m \) glued to it.
• Charge: A total charge \( Q \) is sprinkled uniformly over the surface of the assembly.
• Mechanism: The charges on the surface repel each other. This mutual repulsion creates an outward electrostatic pressure. When the glue fails, this pressure exerts a net force on the disc, causing it to accelerate.
• Approximation: Since \( r \ll R \), we can treat the disc as a small patch on the surface of the sphere.

2. Calculation of Surface Charge Density

The total charge \( Q \) is distributed uniformly over the surface area of the sphere (ignoring the negligible difference in area due to the small flat disc).

Surface Area \( A \approx 4\pi R^2 \).

Surface charge density \( \sigma \) is given by:

$$ \sigma = \frac{Q}{4\pi R^2} $$

3. Force Calculation

The electrostatic pressure \( P \) (force per unit area) acting on a charged surface layer is given by:

$$ P = \frac{\sigma^2}{2\epsilon_0} $$

This pressure acts outwardly, normal to the surface. The force \( F \) acting on the small disc of area \( \pi r^2 \) is:

$$ F = P \times \text{Area of disc} = \frac{\sigma^2}{2\epsilon_0} \times (\pi r^2) $$

Substitute the expression for \( \sigma \):

$$ F = \frac{\pi r^2}{2\epsilon_0} \left( \frac{Q}{4\pi R^2} \right)^2 $$ $$ F = \frac{\pi r^2}{2\epsilon_0} \cdot \frac{Q^2}{16\pi^2 R^4} $$ $$ F = \frac{Q^2 r^2}{32 \pi \epsilon_0 R^4} $$

4. Determination of Acceleration

According to Newton’s second law, \( F = ma \). Therefore, the acceleration \( a \) is:

$$ a = \frac{F}{m} $$ $$ a = \frac{1}{m} \left( \frac{Q^2 r^2}{32 \pi \epsilon_0 R^4} \right) $$